Angular Momentum of a conical pendulum.

AI Thread Summary
The discussion revolves around calculating the angular momentum of a conical pendulum during half a revolution. The initial and final velocities and positions of the ball are defined, leading to the calculation of initial and final angular momentum vectors. A sign error in the initial momentum calculation was identified and corrected, resulting in the correct expression for the change in angular momentum. The conversation also touches on the conservation of angular momentum and the net forces acting on the mass. The participants emphasize the importance of accurately defining coordinate systems and vector components in these calculations.
Satvik Pandey
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Homework Statement


A small ball of mass m suspended from a ceiling at a point O by a thread of length l moves along a horizontal circle with constant angular velocity ##\omega##. Find the magnitude of increment of the vector of the ball's angular momentum relative to point O picked up during half of revolution.

Homework Equations

The Attempt at a Solution



pun.png

I[/B]nitial velocity of ball ##V_{i}=v\hat { j }##

Initial distance of the ball from O is (R)=##lsin\alpha\hat{i}-lcos\alpha\hat{k}##

Final velocity ##V_{f}=-V\hat{j}##

Final distance of the ball from O is ##-lsin\alpha\hat{i}-lcos\alpha\hat{k}##

Initial momentum is ##R## cross ##P##.

##L_{i}=mvLcos\alpha\hat{i}-mvlsin\alpha\hat{k}##

##L_{f}=-mvLcos\alpha\hat{i}+mvlsin\alpha\hat{k}##

##\delta L=-2mvLcos\alpha\hat{i}+2mvlsin\alpha\hat{k}##

So its magnitude is ##2mvL##

I got ##cos\alpha=\frac{g}{\omega^2l}## by writing the force equation.

Now ##v=lsin\alpha\omega##

Using this I got the answer

##2ml^{2}\omega\sqrt{1-\frac{g^{2}}{(\omega^{2}l)^2}}##

But the answer is incorrect.
 
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Can you guys tell me some applications like Daum Equation Editor for writing equations in LaTex. I don't why is it not working in my PC.:mad::confused:
 
Check your coordinate system and the components of your vectors, they do not match o0)
 
ehild said:
Check your coordinate system and the components of your vectors, they do not match o0)

I have considered the circle to be in X-Y plane.
pun.png


I have shown the top view also.

For finding angular momentum of anybody about any point(say O) we drop any line from O to the line of the velocity of that body. Then we calculate ##\overrightarrow { R } \times \overrightarrow { p } ##. right?

So ##R_{i}=lsin(\alpha)\hat { i }- lcos(\alpha)\hat { k }##
 
Satvik Pandey said:
I have considered the circle to be in X-Y plane.View attachment 77622

I have shown the top view also.

For finding angular momentum of anybody about any point(say O) we drop any line from O to the line of the velocity of that body. Then we calculate ##\overrightarrow { R } \times \overrightarrow { p } ##. right?

So ##R_{i}=lsin(\alpha)\hat { i }- lcos(\alpha)\hat { k }##

OK, but then the change the top right figure with the unit vectors. And check the signs in Li .
I got

##L_{i}=mvLcos\alpha\hat{i}+mvlsin\alpha\hat{k}##
 
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ehild said:
OK, but then the change the top right figure with the unit vectors. And check the signs in Li .
I got

##L_{i}=mvLcos\alpha\hat{i}+mvlsin\alpha\hat{k}##

Oh! I made a sign mistake.:mad: Now I got answer

##\frac { 2mgl }{ \omega } \sqrt { 1-{ \left( \frac { g }{ { \omega }^{ 2 }l } \right) }^{ 2 } } ##.

Thank you for the help.:)
 
How is angular momentum conserved from the center of the circle
 
Poojan said:
How is angular momentum conserved from the center of the circle
You mean, about the centre of the circle?
What is the net force on the mass? Where does it point?
 
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